Mathematics Questions and Answers

Number of students who scored atleast 25% = 5 + 3 + 8 = 16

Let A = \(\begin {bmatrix} a & b\\ c & d \end {bmatrix}\)

i.e \(\begin{bmatrix} a & b\\c & d \end{bmatrix}\) \(\begin{bmatrix} 0 & 1\\2 & -1 \end{bmatrix}\) = \(\begin{bmatrix} 2 & -1\\1 & 0 \end{bmatrix}\)

\(\implies \begin {bmatrix} a(0) + b(2) & a(1) + b(-1)\\ c(0) + d(2) & c(1) + d(-1)\end {bmatrix}\) = \(\begin {bmatrix} 2 & -1\\ 1 & 0 \end {bmatrix}\)

\(\implies \begin {bmatrix} 2b & a - b\\2d & c - d\end {bmatrix}\) = \(\begin {bmatrix} 2 & -1\\ 1 & 0 \end {bmatrix}\)

By comparing

2b = 2

a - b = -1

2d = 1 and

c - d = 0

∴ b = \(^2/_2\) = 1

a - b = -1

⇒ a - 1 = -1

∴ a = 0

∴ d = \(^1/_2\)

⇒ c = d

∴ c = \(^1/_2\)

∴The matrice A = \(\begin {bmatrix} 0 & 1\\^1/_2 & ^1/_2 \end {bmatrix}\)

\(\frac {N + 1}{2} = \frac {107 + 1}{2}\) th = 54th value i.e the median age is in the interval "7 - 8" (29 + 25 = 54)

Median = \(1_m + (\frac {^{\sum f} /_2 - f_b}{f_m})\)c

\(l_m\)=lower class boundary of the median age = 6.5

\(\sum f\) = 107

\(f_b\) = sum of all frequencies before the median age = 29

\(f_m\)=frequency of the median age = 40

c = class width = 8.5 - 6.5 = 2

Median = 6.5 + \((\frac {^{107}/_2 - 29}{40})\) x 2

= 6.5 + \((\frac {53.5 - 29}{40})\) x 2

= 6.5 + \((\frac {24.5}{40})\) x 2

= 6.5 + 1.225

\(\therefore\) median age = 7.725

Let the number of white balls be n

The number of red balls = 8

Now, the probability of drawing a white ball = \(\frac {n}{n + 8}\)

The probability of drawing a red ball = \(\frac {8}{n + 8}\)

Since Pr(White ball) = \(\frac{1}{2}\) x Pr(Red ball)

\(\therefore \frac {n}{n + 8} = \frac {1}{2}\times \frac {8}{n + 8}\)

= \(\frac {n}{n + 8} = \frac {4}{n + 8}\)

\(\therefore n = 4\)

Number of white balls = 4

The possible number of outcomes = n + 8 = 4 + 8 = 12

Pr(Red ball and White ball) = Pr(Red ball) x Pr(White ball)

Pr(Red ball) = \(\frac {8}{12}\)

Pr(Red ball) = \(\frac {4}{11}\)

Pr(Red ball and white ball) = \(\frac {8}{12} \times \frac {4}{11}\)

\(\therefore\)Pr(Red ball and white ball) = \(\frac {8}{33}\)

A = { 1, 2, 3, 4, 5, 6}

B = { 2, 4, 6, 8 }

A - B = { 1, 3, 5 }

B - A = { 8 }

∴ (A - B) ⋃ (B - A) = {1, 3, 5, 8}